3
$\begingroup$

Many posts have shown the convergence of the ratio of two consecutive terms in the Fibonacci sequence $F_{n+1} = F_n + F_{n-1}$ when the starting values are $F_0 = 0$ and $F_1 = 1$. How do we show that this holds for any starting values that are integers $a$ and $b$ (at least one is nonzero). That is, for large $n$:

$$ \lim_{n \to \infty} \dfrac{F_{n+1}}{F_n} =\dfrac{1+\sqrt{5}}{2} = \Phi$$

Does that hold for any $a,b \in \mathbb{R}$ provided not both are zero?

$\endgroup$
3
$\begingroup$

When you solve the recurrence $a_n=a_{n-1}+a_{n-2}$, you find that the solutions all have the form

$$a_n=A\varphi^n+B\hat\varphi^n\,,$$

where $\varphi=\frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$; the values of $A$ and $B$ are determined by the initial values $a_0$ and $a_1$. And $\varphi\hat\varphi=-1$, so,

$$\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{A\varphi^{n+1}+B\hat\varphi^{n+1}}{A\varphi^n+B\hat\varphi^n}\\\\ &=\frac{\varphi+\frac{B}A(-1)^n\hat\varphi^{2n+1}}{1+\frac{B}A(-1)^n\hat\varphi^{2n}}\,, \end{align*}$$

and

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\varphi+\frac{B}A(-1)^n\hat\varphi^{2n+1}}{1+\frac{B}A(-1)^n\hat\varphi^{2n}}=\varphi$$

as long as $A\ne0$. If $A=0$, however,

$$\frac{a_{n+1}}{a_n}=\hat\varphi=-\frac1\varphi\,.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.